150 E. Ballico
locally free if rankF rankE , while G = E if rankF = rankE . R
EMARK
1. Let E , h be an A-sheaf, F an A-subsheaf of E and G the saturation of F in E . G is h A-invariant and hence it is an A-sheaf. Since h A, F = h A, G, we have
λ
A
F ≤ λ
A
G, ǫ
A
F ≤ ǫ
A
G, λ
A
F = λ
A
G if and only if G = F and ǫ
A
F = ǫ
A
G if and only if G = F
For any vector bundle F and any line bundle L we have EndF ∼ = EndF ⊗ L and
µ F ⊗ L = µF + degL. This shows that in general the notions of λ
A
-stability, λ
A
- semistability, ǫ
A
-stability and ǫ
A
-semistability are NOT invariant for the twist by a line bundle see Example 1. We believe that ǫ
A
-stability is the correct notion for the Brill - Noether theory
of non-simple vector bundles. In section 3 we will describe all the K-algebras arising for rank two vector bundles.
2. Proofs of Theorems 1, 2 and 3
Let E , h be an A-sheaf on X . Since the saturation of an A-subsheaf of E is an A-subsheaf of E , the usual proof of the existence of an Harder - Narasimhan filtration of any vector bundle on
X see for instance [2], pp. 15–16 gives the following result. P
ROPOSITION
1. Let E , h be an A-sheaf. There is an increasing filtration {E
i
}
0≤i≤r
of E by saturated A-subsheaves such that E = {0}, E
r
= E , E
i
is saturated in E
i+1
for 0 ≤ i r and E
i+1
E
i
is A
i
-semistable, where A
i
⊆ H X, EndE
i+1
E
i
is the image of hE in H
X, EndE
i+1
E
i
and µE
i+1
E
i
µ B for every other A
i
-subsheaf of E E
i
. Proof of Theorem 1. If E is semistable, then obviously it is A-semistable. Assume that E is
not semistable and let F be the first step of the Harder - Narasimhan filtration of E . Thus {0} 6= F and µF µE . By the uniqueness of the Harder - Narasimhan filtration of E
the subsheaf F of E is invariant for the action of AutE . Since AutE is a non-empty open subset of H
X, EndE , F is invariant for the action of the K-algebra H X, EndE . Since
h A ⊆ H X, EndE , F is an A-subsheaf of E . Thus E is not A-semistable.
Proof of Theorem 2. The if part is easy see Example 2. Here we will check the other impli- cation. Since E is polystable, there is an integer s ≥ 1, stable bundles F
1
, . . . , F
s
uniquely determined up to a permutation of their indices with F
i
≇ F
j
if i 6= j and positive integers r
1
, . . . , r
s
such that E ∼ = ⊕
1≤i≤s
F
L r
i
i
. Since E is polystable, µF
i
= µ F
j
for all i, j . Since F
i
and F
j
are stable, with the same slope and not isomorphic, h X, HomF
i
, F
j
= if i 6= j . Hence H
X, EndE ∼ =
L
1≤i≤s
M
r
i
×r
i
K. Since each factor F
⊕r
i
i
is invariant for the action of the group AutE , it is H
X, EndE -invariant and hence h A-invariant, i.e. it is an A-sheaf. Since µF
i
= µ F
j
for any i, j , E is A-stable only if s = 1. Obviously, A is
a unitary K-subalgebra of the unitary K-algebra M
r
1
×r
1
K of r
1
× r
1
matrices and the induced action of A is irreducible because no proper direct factor of F
⊕r
1
1
is A-invariant. Proof of Theorem 3. Since E is semistable but not polystable, the existence of a Jordan - H¨older
filtration of E shows the existence of a maximal proper subsheaf F of E with 0 6= F 6= E and µ
F = µE . Indeed, F contains all proper subsheaves of E with slope µE . Thus F is
Non-simple vector bundles 151
invariant for the action of the group AutE . Hence F is H X, EndE -invariant and hence an
A-sheaf. Thus E is not A-stable. E
XAMPLE
1. Take E , h with A 6= K, rankE = 2 and E non-split extension of a
